Clustering and classification

date()

## [1] "Tue Nov 17 22:47:17 2020"

Data

In this chapter the Boston data from the MASS package is analyzed.

Let’s load and explore the data.

# Accessing the MASS package
library(MASS)

# Loading the Boston data
data("Boston")


# Looking at the structure of the dataset
str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

dim (Boston)

## [1] 506  14

This dataframe contains 506 observation of 14 different variables. These variables are:

crim: per capita crime rate by town.

zn: proportion of residential land zoned for lots over 25,000 sq.ft.

indus: proportion of non-retail business acres per town.

chas: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).

nox: nitrogen oxides concentration (parts per 10 million).

rm: average number of rooms per dwelling.

age: proportion of owner-occupied units built prior to 1940.

dis: weighted mean of distances to five Boston employment centres.

rad: index of accessibility to radial highways.

tax: full-value property-tax rate per $10,000.

ptratio: pupil-teacher ratio by town.

black: 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town.

medv: median value of owner-occupied homes in $1000s.

lstat: lower status of the population (percent).

Graphical overview of the data

Let’s look at the summaries of the variables in the data.

# Looking at summary statistics
summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

All of the variables have different scales. For example, rm varies from 3.5 to 8.7 and tax can vary between 187 to 711. This show that data will need to be standardized when applying clustering as features in the data set have large differences in their ranges. Features with higher ranges would have bigger influence on clustering compared to features with smaller ranges as clustering models are distance based algorithms.

Let’s look at the pairwise relationships between the variables.

# Plotting a matrix of the variables
pairs(Boston)

There are some visible linear relationships between the variables. For example, when medv (median value of owner-occupied homes in $1000s) increases rm (average number of rooms per dwelling) seem to increase as well.

Let’s look at the correlations.

# Accessing corrplot and tidyr libraries 
library(corrplot)

## Warning: package 'corrplot' was built under R version 4.0.3

## corrplot 0.84 loaded

library(tidyr)

## Warning: package 'tidyr' was built under R version 4.0.3

# Loading package for color palletes
library(wesanderson)

## Warning: package 'wesanderson' was built under R version 4.0.3

# Calculating the correlation matrix and rounding it
cor_matrix <- cor(Boston) %>% round(digits = 2)

# Printing the correlation matrix
cor_matrix

##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00

# Visualizing the correlation matrix
corrplot.mixed(cor_matrix, upper.col = wes_palette("Rushmore1", 100, type = "continuous"), lower.col = wes_palette("Rushmore1", 100, type = "continuous"), tl.col= 'black', number.cex = .45, tl.pos = "d", tl.cex = 0.6 )

The highest positive correlation (0.91) is between tax and rad variables. The highest negative correlation (-0.77) is between dis and age variables.

Standardizing the data

Let’s standardize the data by subtracting the column means from the corresponding columns and dividing the difference with standard deviation.

# Scaling the variables
boston_scaled <- scale(Boston)

# Printing summaries of the scaled variables
summary(boston_scaled)

##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

As initial data is a matrix, it needs to be changed into a data frame object.

# Change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
class(boston_scaled)

## [1] "data.frame"

Let’s create a categorical variable of the crime rate in the Boston dataset.

# Using quantiles as break points 
break_points <- quantile(boston_scaled$crim)
break_points

##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610

# Creating a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = break_points, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))

Let’s drop the old crime rate variable from the dataset.

# Removing original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# Adding the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

Let’s divide data into train and test sets.

# Setting the number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# Choosing randomly 80% of the rows
set.seed(123)
ind <- sample(n,  size = n * 0.8)

# Creating a train set

train <- boston_scaled[ind,]

# Creating a test set 
test <- boston_scaled[-ind,]

Fitting the linear discriminant analysis

In this section LDA will be fitted using the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.

# Fitting LDA
lda.fit <- lda(crime ~ ., data = train)

# Printing the lda.fit object
lda.fit

## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2549505 0.2500000 0.2500000 0.2450495 
## 
## Group means:
##                   zn      indus        chas        nox          rm        age
## low       1.01866313 -0.9066422 -0.08120770 -0.8835724  0.38666334 -0.9213895
## med_low  -0.07141687 -0.3429155  0.03952046 -0.5768545 -0.09882533 -0.3254849
## med_high -0.40148591  0.2162741  0.19544522  0.3637157  0.12480815  0.4564260
## high     -0.48724019  1.0149946 -0.03371693  1.0481437 -0.47733231  0.8274496
##                 dis        rad        tax    ptratio      black        lstat
## low       0.9094324 -0.6819317 -0.7458486 -0.4234280  0.3729083 -0.766883775
## med_low   0.3694282 -0.5395408 -0.4205644 -0.1079710  0.3103406 -0.164921798
## med_high -0.3720478 -0.4349280 -0.3191090 -0.2716959  0.1049654  0.009720124
## high     -0.8601246  1.6596029  1.5294129  0.8057784 -0.6383964  0.900379309
##                 medv
## low       0.47009410
## med_low   0.01548761
## med_high  0.19440747
## high     -0.71294711
## 
## Coefficients of linear discriminants:
##                  LD1         LD2        LD3
## zn       0.148390645  0.74870532 -1.0874785
## indus    0.040971465 -0.38126652 -0.1619456
## chas     0.002460776  0.03963849  0.1699312
## nox      0.312458150 -0.67564471 -1.3104018
## rm       0.011086947 -0.16058718 -0.1572603
## age      0.283482486 -0.38817624 -0.1013491
## dis     -0.079848789 -0.38493775  0.2108038
## rad      3.718978412  0.93123532 -0.4706522
## tax     -0.015197127  0.04230505  1.2889075
## ptratio  0.180294774 -0.01592588 -0.3558490
## black   -0.136724112  0.02948075  0.1288959
## lstat    0.145739238 -0.37823065  0.3345688
## medv     0.061327205 -0.53906503 -0.1509890
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9523 0.0364 0.0113

# The function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# Setting target classes as numeric
classes <- as.numeric(train$crime)

# Plotting the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)

From LDA bi-plot it can be seen that rad variable is the most infuential linear separator for discriminating high crime rates. Although, some points of medium high rates are mixed in.

Making predictions with LDA model

Let’s save the crime categories from the test set and then remove it from the test dataset.

# Saving the correct classes from test data
correct_classes <- test$crime

# Removing the crime variable from test data
test <- dplyr::select(test, -crime)

NOw let’s make predictions with LDA model on the test set and cross tabulate the results.

# Predicting the classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# Cross tabulating the results
table(correct = correct_classes, predicted = lda.pred$class)

##           predicted
## correct    low med_low med_high high
##   low       10      10        4    0
##   med_low    2      17        6    0
##   med_high   1       9       13    2
##   high       0       0        1   27

It seems that the model best predicts high crime rates. All 12 predictions predicting high crime rate were correct. It is hard for the model to separate between low, medium low and medium high crime rates.

Clustering

Let’s reload and scale Boston data set.

# Scaling the variables in Boston data set
boston_scaled2 <- scale(Boston)

Let’s calculate the Euclidean distances between the variables.

# Distances between the observations
dist_eu <- dist(Boston)

Let’s run k-means algorithm on the dataset choosing 2 clusters.

# k-means clustering
km <-kmeans(Boston, centers = 2)

# Plotting the Boston dataset with clusters
pairs(Boston, col = km$cluster)

Let’s investigate what is the optimal number of clusters and run the algorithm again.

library(ggplot2)

## Warning: package 'ggplot2' was built under R version 4.0.3

# Setting seed so that initial clusters would be assinged every time the same
set.seed(123)

# Determining the number of clusters
k_max <- 20

# Calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# Visualizing the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

Based on first sharp drop it seems that optimal number of clusters is as chosen initially. That is 2 clusters.

Let’s plot the clusters in the data.

library(GGally)

## Warning: package 'GGally' was built under R version 4.0.3

## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2

ggpairs(Boston, mapping = aes(color = factor(km$cluster)))

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

## Warning in cor(x, y): the standard deviation is zero

From the last plot it is visible that one of the clusters contains most high crime points and the other all the rest points as high crime points were well separated with LDA based on rad variable.

Projection of the data points (super bonus)

model_predictors <- dplyr::select(train, -crime)
# Checking the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

# Matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

#install.packages('plotly')
library(plotly)

## Warning: package 'plotly' was built under R version 4.0.3

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
## 
##     last_plot

## The following object is masked from 'package:MASS':
## 
##     select

## The following object is masked from 'package:stats':
## 
##     filter

## The following object is masked from 'package:graphics':
## 
##     layout

# Plotting with color argument equal to crime classes
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.

# Plotting with color argument equal to kmeans clusters of the observations in train set.
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = factor(km$cluster[ind]))

## Warning in RColorBrewer::brewer.pal(N, "Set2"): minimal value for n is 3, returning requested palette with 3 different levels

## Warning in RColorBrewer::brewer.pal(N, "Set2"): minimal value for n is 3, returning requested palette with 3 different levels

These two plots differ in a way that there are only two clusters in the second plot and 4 different classes in the first plot. The similarly here is that high crime points from the first plot are assigned to the first cluster.